# Question Video: Determining the Number of Real Roots of a Quadratic Equation Using the Discriminant

How many real roots does the equation ππ₯Β² + ππ₯ + π = 0 have if a β  0 and πΒ² β 4ππ = 0?

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### Video Transcript

How many real roots does the equation ππ₯ squared plus ππ₯ plus π equals zero have if π is not equal to zero and π squared minus four ππ is equal to zero?

Well, this equation is the general form of a quadratic equation. And hopefully, you can recall that if ππ₯ squared plus ππ₯ plus π equals zero, then π₯ is equal to the negative of π plus or minus the square root of π squared minus four ππ all over two π. Now obviously, this only works if π is not equal to zero. Otherwise, weβd be dividing something through it by zero. And they tell us that π is not equal to zero in the question, so thatβs good. And the other bit of information they give us is that π squared minus four ππ is equal to zero.

Now the value of π squared minus four ππ has a special name; we call it the discriminant. And itβs important because itβs the expression under the square root in this formula. And weβre told that in this particular case, that is equal to zero. So this means that the root of the equation, the possible values of π₯, are the negative value of the π coefficient plus or minus the square root of zero all over two times the value of the π coefficient. Mostly, the square root of zero is zero. So weβve got the negative value of π plus zero or minus zero, thatβs only one possible number, all over two π. So negative π plus zero or negative π minus zero. And whether you add zero or subtract zero, it doesnβt make any difference. You are just gonna get a value of negative π on the numerator, and youβre gonna get two expressions that look the same, negative π over two π in both cases.

So our formula gives rise in its two various forms to just the same one solution. So under the conditions given in the question, the answer is that there is only one real root.